Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $k \neq 0$. $q = \dfrac{3k}{3(3k + 7)} \div \dfrac{7k}{3k + 7} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{3k}{3(3k + 7)} \times \dfrac{3k + 7}{7k} $ When multiplying fractions, we multiply the numerators and the denominators. $q = \dfrac{ 3k \times (3k + 7) } { 3(3k + 7) \times 7k } $ $ q = \dfrac {3k (3k + 7)} {7k \times 3(3k + 7)} $ $ q = \dfrac{3k(3k + 7)}{21k(3k + 7)} $ We can cancel the $3k + 7$ so long as $3k + 7 \neq 0$ Therefore $k \neq -\dfrac{7}{3}$ $q = \dfrac{3k \cancel{(3k + 7})}{21k \cancel{(3k + 7)}} = \dfrac{3k}{21k} = \dfrac{1}{7} $